1. Chemistry & Stoichiometry living in harmon-ee!

2. Consider this equation for respiration: C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 0 If I have 3 mol of C 6 H 12 O 6, how many mol of CO 2 will I make?

3.  Check you answers with your partner first  I will put up answers in 2 minutes

4.  I can determine how many grams of a product will be made from a given quantity of a reactant using stoichiometry

5.  Mass of CO 2  Moles of CO 2  Moles of X 2 CO 3  Molar Mass of X 2 CO 3  Mass of X  Identity of X Use Molar Mass! Use Mole Ratios! Use Proportions! Subtract out the CO3! Use Mass # on Per. Table!

6.  You have 5 minutes to turn in your lab to the pink basket  If you are finished start on your homework  “Intro to Stoich” Worksheet

7. This is how your car/bus got you to school:  2 C 8 H 18 + 25 O 2 → 16 CO 2 + 18 H 2 O  If you have 5.5 mol of C 8 H 18, how many mol of CO 2 will your car produce? Discuss with your table partner how to set up the problem. I’ll call on someone to guide me through it.

8. 2 C 8 H 18 + 25 O 2 → 16 CO 2 + 18 H 2 O  If you have 5.5 mol of C 8 H 18, how many mol of CO 2 will your car produce? (44 mol)  If you have 5.5 grams of C 8 H 18, how many grams of CO 2 did you produce?  Use Molar Mass!

9. If you have 5.5 grams of C 8 H 18, how many grams of CO 2 did you produce? Strategy: grams of C 8 H18  mol C 8 H 18  mol CO 2  g CO 2 5.5 g C 8 H 18 x 1 mol C 8 H 18 x 16 mol CO 2 x 44.01 g CO 2 114 g C 8 H 18 2 mol C 8 H 18 1 mol CO 2 = 16.98631579 = 17 g CO 2 (2 sig figs)

10. What you started with 8.00g of H 2, how many grams of CH 3 OH do you produce? Strategy?? g of H 2  mol H 2  mol CH 3 OH  g of CH 3 OH 8.00 g H 2 x 1 mol H 2 x 1 mol CH 3 OH x 32.05g CH 3 OH 2.o2g H 2 2 mol H 2 1 mol CH 3 OH = 63.46534653 g CH 3 OH  63.5 g CH 3 OH

11.  Let’s work on #1 together  The rest independently, or with your partner  Skip #4, 6, 7, and 9… we will get to those.

12. Limiting reactants

13. Say you wanted to write everyone letters (because you’re awesome like that). You wanted each letter to have 1 envelope, 1 stamp, and 1 card. You have 1000 envelops, 100 stamps, and 253 cards. How many letters could you write? Why? How did you figure that out? How come you make more?

14.  Limiting Reactant = Reactant that runs out 1st  Limits how much product is produced  When this chemical runs out the reaction stops!!  Leftovers = Excess Reactant(s): Have Extra  Reactant(s) that you have extra of.  Not completely used up in the reaction.  If they aren’t used up, then they never react!

15.  Treat it like 2 separate problems  Solve it with both starting #’s  Whichever makes less is the limiting reactant, since you can only make as much and the least amount!

16. Get a laptop to share with your elbow partner.

17.  Go to the following website: Web link in my public file  http://phet.colorado.edu/en/simulation/reactants- products-and-leftovershttp://phet.colorado.edu/en/simulation/reactants- products-and-leftovers  Click on play button.  Set the coefficients as to how you want to make a cheese sandwich  Then change the amount of reactants, as you do this pay attention to what is happening to the products & leftovers?  Do the same thing for the meat & cheese sandwich?  Answer this question: What is the role of the coefficients? What do the coefficients “measure”? Homework:  Finish PhET  Finish “Intro to Stoich” W.S

18. For the following reaction: CaC 2(s) + 2H 2 O (l)  C 2 H 2(g) + Ca(OH) 2(s) 1. If you react 0.400 g of CaC 2, how many moles of acetylene (C 2 H 2 ) will you produce? 2. What is the name of the other product, Ca(OH) 2(s) ? 3. If 1 mole of acetylene gas = 22.4 L, how many mL of acetylene is that? (Two conversions!)

19. For the following reaction: CaC 2(s) + 2H 2 O (l)  C 2 H 2(g) + Ca(OH) 2(s) 1. If you react 0.400 g of CaC2, how many moles of acetylene (C 2 H 2 ) will you produce?

20. For the following reaction: CaC 2(s) + 2H 2 O (l)  C 2 H 2(g) + Ca(OH) 2(s) 3. If 1 mole of acetylene gas = 22.4 L, how many mL of acetylene is that? (Two conversions!)

21. 0.4 g CaC2  139 mL C2H2 **Needs 347.5 mL O2 to fully react. Total volume: 486.5 mL 4 g CaC2  1390 mL C2H2 **Needs 3475 mL O2 to fully react. Total volume: 4865 mL 20 oz. bottle = 591 mL bottle.